#1**0 **

Since there are only 32 numbers to try: 0^{2}, 1^{2}, 2^{2}, 3^{2}, ..., 31^{2} (32^{2} is larger than 1000), it would be easiest just to try them all.

However, if you want a little more orgainized way, you might consider this:

What is the result when the

-- one's digit is a 1? 1

-- one's digit is a 2? 4 x

-- one's digit is a 3? 9

-- one's digit is a 4? 6

-- one's digit is a 5? 5

-- one's digit is a 6? 6

-- one's digit is a 7? 9

-- one's digit is an 8? 4 x

-- one's digit is a 9? 1

So the question becomes: How many numbers smaller than 32 have a one's digit which is either a "2" or an "8"?

geno3141 May 30, 2020